Question

How do you evaluate the integral `int sqrt(4+x^2)`?

Answer 1

`I = x sqrt(1 + x^2/4) + 2 sinh^(-1) (x/2) + C`

Explanation:

`I = int sqrt(4+x^2) color(red)(dx)`

You can use identity:

• `cosh^2 y - sinh^2 y = 1 implies cosh^2y = 1 + sinh^2 y`

So let:

• `x^2 = 4 sinh^2 y`

• `implies 2 x \ dx = 8 sinh y \ cosh y \ dy`

`implies I = int sqrt(4+4 sinh^2 y) \ (8 sinh y \ cosh y \ dy)/(2x)`

`= int 2 cosh y \ (8 sinh y \ cosh y \ dy)/(2* 2 sinh y)`

`=4 int cosh^2 y \ dy`

`=4 int (cosh 2y +1 ) /2\ dy`

`= sinh 2y + 2 y + C qquad triangle`

Considering:

• `color(red)(sinh 2y = 2 sinh y cosh y )`

• `x^2 = 4 sinh^2 y implies color(red)( x = 2 sinh y) color(red)(implies y = sinh^(-1) (x/2))`

• `cosh^2y = 1 + sinh^2 y implies color(red)(cosh y = sqrt(1 + x^2/4) )`

Then `triangle` becomes:

`I = 2 * x/2 * sqrt(1 + x^2/4) + 2 sinh^(-1) (x/2) + C`

`= x sqrt(1 + x^2/4) + 2 sinh^(-1) (x/2) + C`